log ex is equal to:

Question

log ex is equal to: (A)  (x-1)-((x-1)2/2)+((x-1)3/3)-....∞  (B) 0 (C) 1 (D) none of the above

Tardigrade Admin · Accepted Answer

The series of log, (1+x) is x-(x2/2)+(x2/3)-(x4/4)+.... Now, log ex= log e(1+(x-1)) =(x-1)-((x-1)2/2)+((x-1)3/3)-.... Note: the series of log (1+x) is defined only when -1 < x < 1.

Q. $lo g_{e} x$ is equal to:

$(x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - ....\infty$

0

1

none of the above

The series of log, $(1 + x)$ is
$x - \frac{x ^{2}}{2} + \frac{x ^{2}}{3} - \frac{x ^{4}}{4} + ....$
Now, $lo g_{e} x = lo g_{e} (1 + (x - 1))$
$= (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - ....$
Note: the series of $lo g (1 + x)$ is defined only when $- 1 < x < 1.$

Q. loge​x is equal to:

(x−1)−2(x−1)2​+3(x−1)3​−....∞

0

1

none of the above

The series of log, (1+x) is x−2x2​+3x2​−4x4​+.... Now, loge​x=loge​(1+(x−1)) =(x−1)−2(x−1)2​+3(x−1)3​−.... Note: the series of log(1+x) is defined only when −1<x<1.

Q. $lo g_{e} x$ is equal to:

$(x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - ....\infty$

The series of log, $(1 + x)$ is
$x - \frac{x ^{2}}{2} + \frac{x ^{2}}{3} - \frac{x ^{4}}{4} + ....$
Now, $lo g_{e} x = lo g_{e} (1 + (x - 1))$
$= (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - ....$
Note: the series of $lo g (1 + x)$ is defined only when $- 1 < x < 1.$